= 1 and so it must be that C = 1. Therefore, there is exactly one solution to
the initial value problem in 8.14 and it is y

(x)

=

(1+ x)

α.

The strategy for finding the Taylor series of this function consists of finding a series which
solves the initial value problem above. Let

∞∑
y(x) ≡ anxn (8.16)
n=0

(8.16)

be a solution to 8.14. Of course it is not known at this time whether such a series exists.
However, the process of finding it will demonstrate its existence. From Theorem 8.2.1 and the
initial value problem,

∞∑ n−1 ∑∞ n
(1 +x ) annx − αanx = 0
n=0 n=0

and so

∞∑ n−1 ∞∑ n
annx + an(n− α) x = 0
n=1 n=0

Changing the variable of summation in the first sum,

∑∞ ∑∞
an+1(n + 1)xn + an (n − α)xn = 0
n=0 n=0

and from Corollary 8.2.2 and the initial condition for 8.14 this requires

a (α− n)
an+1 = -n------,a0 = 1. (8.17)
n+ 1

(8.17)

Therefore, from 8.17 and letting n = 0, a1 = α, then using 8.17 again along with this
information,

a2 = α-(α-−-1).
2

Using the same process,

( α(α−1))
----2----(α-−-2) α(α−-1)(α-−-2)
a3 = 3 = 3! .

By now you can spot the pattern. In general,

◜---n-of thes◞e◟ factors-◝
α-(α-−-1)⋅⋅⋅(α-−-n+-1)
an = n! .

Therefore, the candidate for the Taylor series is

∑∞ α-(α−-1)⋅⋅⋅(α−-n-+-1) n
y(x) = n! x .
n=0

Furthermore, the above discussion shows this series solves the initial value problem on its
interval of convergence. It only remains to show the radius of convergence of this series equals
1. It will then follow that this series equals

(1+ x)

α because of uniqueness of the
initial value problem. To find the radius of convergence, use the ratio test. Thus the
ratio of the absolute values of

showing that the radius of convergence is 1 since the series converges if

|x|

< 1 and diverges if

|x|

> 1.

The expression,

α(α− 1)⋅⋅⋅(α−n+1)
------n!-----

is often denoted as

(α)
n

. With this notation, the following
theorem has been established.

Theorem 8.4.2Let α be a real numberand let

|x|

< 1. Then

∞∑ ( )
(1+ x)α = α xn.
n=0 n

There is a very interesting issue related to the above theorem which illustrates the
limitation of power series. The function f

(x)

=

(1 + x)

α makes sense for all x > −1 but one
is only able to describe it with a power series on the interval

(− 1,1)

. Think about this. The
above technique is a standard one for obtaining solutions of differential equations and this
example illustrates a deficiency in the method.

To completely understand power series, it is necessary to take a course in complex
analysis. It turns out that the right way to consider Taylor series is through the use of
geometric series and something called the Cauchy integral formula of complex analysis.
However, these are topics for another course.